Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.Scanner;

public class Main {
    static Scanner sc = new Scanner(System.in);

    public static void main(String[] args) {
        int N = sc.nextInt();
        double[] p = new double[N + 1];
        double[][] f = new double[N + 1][N + 1];
        for(int i = 1; i <= N;  i++)p[i] = sc.nextDouble();

        f[0][0] = 1;
        for(int i = 1; i <= N; i ++){
            f[i][0] = f[i - 1][0] * (1 - p[i]);
            for(int j = 1; j <= i; j ++){
                f[i][j] = f[i - 1][j - 1] * p[i] + f[i - 1][j] * (1 - p[i]);
            }
        }
        int cnt;
        if((N&1) == 1) cnt = N + 1 >> 1;
        else cnt = (N >> 1) + 1;
        double res = 0.0;
        for(int i = cnt; i <= N; i++)res += f[N][i];
        System.out.println(res);
    }
}